Normality — Definition, Formula, and Relation to Molarity

Definition: Normality (N) = Number of gram equivalents of solute / Volume of solution in litres

Formula: N = (Weight of solute / Equivalent weight of solute) / Volume of solution (in litres)

OR: N = Molarity (M) × n-factor

Where:

• Equivalent weight = Molecular weight / n-factor
• n-factor depends on the reaction type: • For acids: n-factor = number of H⁺ ions (basicity) • For bases: n-factor = number of OH⁻ ions (acidity) • For oxidising/reducing agents: n-factor = change in oxidation number × number of atoms

Unit: N (or eq/L, equivalents per litre)

Acids:

• HCl: n = 1 (releases 1 H⁺)
• H₂SO₄: n = 2 (releases 2 H⁺)
• H₃PO₄: n = 3 (releases 3 H⁺)

Bases:

• NaOH: n = 1 (releases 1 OH⁻)
• Ca(OH)₂: n = 2 (releases 2 OH⁻)
• Al(OH)₃: n = 3 (releases 3 OH⁻)

Salts (in acid-base context):

• Na₂CO₃: n = 2
• NaHCO₃: n = 1

Oxidising agents:

• KMnO₄ in acidic medium: n = 5 (Mn⁷⁺ → Mn²⁺)
• KMnO₄ in neutral/alkaline: n = 3 (Mn⁷⁺ → Mn⁴⁺)
• K₂Cr₂O₇: n = 6

Example 1: Find normality of 9.8 g of H₂SO₄ in 500 mL solution.

• Molecular weight of H₂SO₄ = 98 g/mol
• n-factor (basicity) = 2
• Equivalent weight = 98/2 = 49 g/eq
• Number of equivalents = 9.8/49 = 0.2 eq
• Volume = 500 mL = 0.5 L
• N = 0.2/0.5 = 0.4 N

Example 2: 1 M H₂SO₄ → Normality = ?

• N = M × n-factor = 1 × 2 = 2 N

Example 3: 1 M NaOH → Normality = ?

• N = 1 × 1 = 1 N (n-factor of NaOH = 1)

Molarity (M):

• Moles of solute per litre
• M = moles/litre
• Independent of reaction type

Normality (N):

• Gram equivalents per litre
• N = M × n-factor
• Depends on the reaction

Key relationship: N = M × n-factor

For compounds where n-factor = 1 (e.g., HCl, NaOH): N = M For compounds where n-factor = 2 (e.g., H₂SO₄): N = 2M

Important: Normality of a solution is not fixed — it changes with the reaction context. For example, H₃PO₄ can have N = M, N = 2M, or N = 3M depending on how many protons it donates in a reaction.